What is the integral of 7(7x+4)^5 ?

The integral of 7(7x+4)^5 is 1/6 (7x+4)^6+C

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integral of 7(7x+4)^5

\int 7 (7 x + 4)^{5} d x

\frac{1}{6} (7 x + 4)^{6} + C

Solution steps

\int 7 (7 x + 4)^{5} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 7 \cdot \int (7 x + 4)^{5} d x

Apply u-substitution

: \int \frac{u ^{5}}{7} d u

= 7 \cdot \int \frac{u ^{5}}{7} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 7 \cdot \frac{1}{7} \cdot \int u^{5} d u

Apply the Power Rule

: \frac{u ^{6}}{6}

= 7 \cdot \frac{1}{7} \cdot \frac{u ^{6}}{6}

Substitute back

u = 7 x + 4

= 7 \cdot \frac{1}{7} \cdot \frac{( 7 x + 4 ) ^{6}}{6}

Simplify

7 \cdot \frac{1}{7} \cdot \frac{( 7 x + 4 ) ^{6}}{6} : \frac{1}{6} (7 x + 4)^{6}

= \frac{1}{6} (7 x + 4)^{6}

Add a constant to the solution

= \frac{1}{6} (7 x + 4)^{6} + C

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